To solve for $a$, let's analyze the given information in the image.

Given:

1. $Z_1 = a + 3i$
2. $Z_2 = -2 - 3i$
3. $Z_1 Z_2 = 7 - 9i$

We need to find the value of $a$.

Solution:

We know that: $Z_1 Z_2 = (a + 3i)(-2 - 3i)$

Expanding $Z_1 Z_2$:

Using the distributive property: $Z_1 Z_2 = a(-2 - 3i) + 3i(-2 - 3i)$

Breaking it down:

1. $a \cdot (-2) = -2a$
2. $a \cdot (-3i) = -3ai$
3. $3i \cdot (-2) = -6i$
4. $3i \cdot (-3i) = -9i^2$

Since $i^2 = -1$, we substitute this in the last term: $3i \cdot (-3i) = -9(-1) = 9$

Now, combine all terms: $Z_1 Z_2 = (-2a + 9) + (-3a - 6)i$

Equating with $Z_1 Z_2 = 7 - 9i$:

From the real parts: $-2a + 9 = 7$

From the imaginary parts: $-3a - 6 = -9$

Solving these equations:

1. Real part equation: $-2a + 9 = 7$ $-2a = 7 - 9$ $-2a = -2$ $a = 1$

2. Imaginary part equation (for verification): $-3a - 6 = -9$ $-3a = -9 + 6$ $-3a = -3$ $a = 1$

Both parts confirm that $a = 1$.

Answer:

$a = 1$

Would you like further details on this solution, or have additional questions?

Related Questions:

1. How would you verify the answer by substituting back into the equation?
2. What is the general form for multiplying two complex numbers?
3. How do the imaginary and real parts separately affect the solution?
4. Can complex numbers with imaginary components equal real numbers?
5. How is this method useful in real-world applications of complex numbers?

Tip: When multiplying complex numbers, always simplify using $i^2 = -1$ to handle imaginary units effectively.